Respuesta :
Hello,
function minmax(int p1,int p2,int p3, int adr_big, int adr_small)
{ int mini=p1,maxi=p1;
if (p1>p2) {mini=p2;}
else {maxi=p2;};
if (p3>maxi) maxi=p3;
if (p3<mini) mini=p3;
*adr_big=maxi;
*adr_small=mini;
};
// main
int a=31,b=5,c=19,big,small;
minmax(a,b,c,&big,&small);
function minmax(int p1,int p2,int p3, int adr_big, int adr_small)
{ int mini=p1,maxi=p1;
if (p1>p2) {mini=p2;}
else {maxi=p2;};
if (p3>maxi) maxi=p3;
if (p3<mini) mini=p3;
*adr_big=maxi;
*adr_small=mini;
};
// main
int a=31,b=5,c=19,big,small;
minmax(a,b,c,&big,&small);
Answer:
void minMax(int a, int b, int c, int*big, int*small)
{
if(a>b && a >c){
*big = a;
if(b>c)
*small = c;
else
*small = b;
}
else if (b>a && b>c){
*big = b;
if(a>c)
*small = c;
else
*small = a;
}
else{
*big = c;
if(a>b)
*small = b;
else
*small = a;
}
}
Step-by-step explanation:
MPL